A certain string can withstand a maximum tensiion of 40 N without breaking. A child ties a 0.37 kg stone to one end and, holding the other end, whirls the stone in a vertical circle of radius 0.91 m, slowly increasing the speed until the string breaks. (a) Where is the stone on its path when the string breaks? (b) What is the speed of the stone as the string breaks?

Answer

To solve the problem step by step, we will analyze the forces acting on the stone as it is whirled in a vertical circle and determine the conditions under which the string will break.

### Step 1: Identify the Forces Acting on the Stone
When the stone is at the lowest point of the circular path, two forces act on it:
1. The gravitational force acting downwards, \( F_g = mg \)
2. The tension in the string acting upwards, \( T \)

### Step 2: Apply Newton's Second Law
At the lowest point of the circular motion, the net force acting on the stone provides the centripetal force required for circular motion. According to Newton's second law, we can write:
\[ T - mg = \frac{mv^2}{r} \]
Where:
- \( T \) is the tension in the string,
- \( m \) is the mass of the stone (0.37 kg),
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)),
- \( v \) is the speed of the stone,
- \( r \) is the radius of the circle (0.91 m).

### Step 3: Set Up the Equation for Maximum Tension
The string can withstand a maximum tension of 40 N before breaking. Therefore, we set \( T = 40 \, \text{N} \):
\[ 40 - mg = \frac{mv^2}{r} \]

### Step 4: Substitute Known Values
Substituting the known values into the equation:
\[ 40 - (0.37 \times 9.81) = \frac{0.37 v^2}{0.91} \]

Calculating \( mg \):
\[ mg = 0.37 \times 9.81 \approx 3.6277 \, \text{N} \]

Now substituting this back into the equation:
\[ 40 - 3.6277 = \frac{0.37 v^2}{0.91} \]
\[ 36.3723 = \frac{0.37 v^2}{0.91} \]

### Step 5: Solve for \( v^2 \)
Rearranging the equation to solve for \( v^2 \):
\[ v^2 = \frac{36.3723 \times 0.91}{0.37} \]
Calculating the right side:
\[ v^2 \approx \frac{33.0739}{0.37} \approx 89.1 \]

### Step 6: Calculate \( v \)
Taking the square root to find \( v \):
\[ v \approx \sqrt{89.1} \approx 9.43 \, \text{m/s} \]

### Step 7: Determine the Position of the Stone When the String Breaks
The string will break when the stone is at the lowest point of the circular path because this is when the tension is maximized.

### Final Answers
(a) The stone is at the **lowest point** of its path when the string breaks.
(b) The speed of the stone as the string breaks is approximately **9.43 m/s**.